SOLUTION: conjecture a formula for the following product the product of the nth number when (1-(4/(2i-1)^2))= (1-(4/(1)^2))(1-(4/(3)^2))(1-(4/(5)^2))...(1-(4/(2n-1)^2)).

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Question 28682: conjecture a formula for the following product
the product of the nth number when (1-(4/(2i-1)^2))= (1-(4/(1)^2))(1-(4/(3)^2))(1-(4/(5)^2))...(1-(4/(2n-1)^2)).
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
onjecture a formula for the following product
the product of the nth number when (1-(4/(2i-1)^2))=
LET PN =(1-(4/(1)^2))(1-(4/(3)^2))(1-(4/(5)^2))...(1-(4/(2n-1)^2)),when i=1.
WHAT DO YOU MEAN BY I=1...I THINK IT SHOULD BE JUST AS GIVEN IN RHS WITH I=N...FURTHER USE OF I IS NOT DESIRABLE IN THESE SUMS AS IT MAY BE MISTAKEN AS THE COMPLEX/IMAGINARY NUMBER I.ANY WAY LET US DO THE PROBLEM BY TACKLING THE GENERAL FORM ...............(1-(4/(2n-1)^2))
={((2N-1)^2-4)/((2n-1)^2)}
=(4N^2-4N+1-4)/(2n-1)^2
=(4N^2-4N-3)/(2N-1)^2
=(4N^2-6N+2N-3)/(2N-1)^2
={2N(2N-3)+1(2N-3)}/(2N-1)^2
=(2N-3)(2N+1)/(2N-1)^2
HENCE THE REQUIRED PRODUCT PN IS OBTAINED BY PUTTING N=1,2,3....ETC..TILL N IN THIS.
PN= ((-1)*3/1^2)(1*5/3^2)(3*7/5^2)..{(2N-5)(2N-1)/(2N-3)^2}{(2N-3)(2N+1)/(2N-1)^2}
WE FIND ALL EXCEPT THE FOLLOWING CANCEL OUT..SO
PN=(-1)(2N+1)/(2N-1)^2